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Sphinxa [80]
3 years ago
11

A chemist adds 255 .0 mL of a M copper(II) sulfate solution to a reaction flask. Calculate the mass in grams of copper(II) sulfa

te the chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
melomori [17]3 years ago
7 0

Answer:

The correct answer is 40.7 grams.

Explanation:

Based on the given information, the volume of copper sulfate added to the solution is 255 ml or 0.255 L. The molarity of copper sulfate is not given, so let us consider it to be 1 M, which can also be written as 1 moles per liter.

The moles of copper sulfate can be determined by using the formula,

Moles = Molarity * Volume in Liters

Moles of CuSO4 = 1 moles/Liter * 0.255 L

Moles of CuSO4 = 0.255 moles

The mass of CuSO4 added in the solution will be,

Mass of CuSO4 = Moles * Molecular mass

= 0.255 moles * 159.609 grams per mole

= 40.7 grams.

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The Al2O3 crystal structure (corundum) consists of an HCP arrangement of O2- ions; the Al3 ions occupy octahedral positions. Wha
ddd [48]

Answer:

2/3

Explanation:

Crystals structures can also be seen when two elements combines together and the perfect example is Al₂O₃ which is given in the question above. Just like it is given in the question above, the kind of arrangement in the crystal structure for Al₂O₃ is called HCP which stands for Hexagonally Closed Pack.

The aluminum ions which is in form of Al³⁺ occupies the two-third[2/3] positions while the position that the oxygen ion occupies is one[1].

8 0
3 years ago
A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
OLEGan [10]

Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

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